Gears
10.11
1
PATH OF CONTACT
Let two gear wheels with centres A and B be in contact (Fig. 10.24).
Fig. 10.24
The pinion 1 is the driver and is rotating clockwise. The wheel 2 is driven in
the counter-clockwise direction. EF is their common tangent to the base circles.
Contact of the two teeth is made where the addendum circle of the wheel
meets the line of action EF, i.e. at C and is broken where the addendum circle
of the pinion meets the line of action, i.e. at D. CD is then the path of contact.
Let r = pitch circle radius of pinion
R = pitch circle radius of wheel
ra = addendum circle radius of pinion
Ra = addendum circle radius of wheel.
Path of contact = path of approach + path of recess
CD = CP + PD
= (CF – PF) + (DE – PE)
= ÈÎ Ra2 - R 2 cos 2 j - R sin j ˘˚ + ÈÎ ra2 - r 2 cos 2 j - r sin j ˘˚
2
Theory of Machines
(10.5)
Ra2 - R 2 cos2 j + ra2 - r 2 cos2 j - ( R + r ) sin j
Observe that the path of approach can be found if the dimensions of the
driven wheel are known. Similarly, the path of recess is known from the
dimensions of the driving wheel (pinion).
=
10.12
ARC OF CONTACT
Arc of contact is the distance travelled by a point on either pitch circle of the
two wheels during the period of contact of a pair of teeth.
In Fig.10.25, at the beginning of engagement, the driving involute is shown
as GH; when the point of contact is at P, it is shown as JK and when at the end
of engagement, it is DL. The arc of contact is P¢P¢¢ and it consists of the arc of
approach P¢P and the arc of recess PP¢¢.
Fig. 10.25
Let the time to traverse the arc of approach is ta, then
Arc of approach = P¢P = Tangential velocity of P¢ ¥ Time of approach
(ta = time of approach)
= wa r ¥ ta
= wa (r cos j)
1
t
cos j a
= (Tang. vel. of H) ta
1
cos j
=
Arc HK Arc FK - Arc FH
=
cos j
cos j
=
FP - FC
CP
=
cos j
cos j
(AF = AH)
Arc FK is equal to the path FP as the point P is on the generator FP that
rolls on the base circle FHK to generate the involute PK. Similarly, arc FH =
Path FC.
3
Gears
Arc of recess = PP¢¢ = Tang. vel. of P ¥ Time of recess
(tr = time of recess)
= wa r ¥ tr
= wa (r cos j)
1
t
cos j r
= (Tang. vel. of K) tr
or
=
Arc KL
cos j
=
Arc FL - Arc FK
cos j
PP¢¢ =
1
cos j
FD - FP
PD
=
cos j
cos j
Arc of contact =
CP + PD
CP
PD
CD
+
=
=
cos j cos j
cos j
cos j
or
Arc of contact =
Path of contact
cos j
10.13
NUMBER OF PAIRS OF TEETH IN CONTACT
(10.6)
The arc of contact is the length of the pitch circle traversed by a point on it
during the mating of a pair of teeth.
Thus, all the teeth lying in between the arc of contact P ¢P ¢¢ will be meshing
with the teeth on the other wheel.
Number of teeth within the arc P ¢P ¢¢ =
n=
Arc P ¢P ¢¢
Circular pitch
CD 1
cos j p
(10.7)
For continuous transmission of motion, at least one tooth of one wheel must
be in contact with another tooth of the second wheel. Therefore, n must be
greater than unity.
If n lies between 1 and 2, the number of teeth in contact at any time will not
be less than one and never more than two. If n is between 2 and 3, it is never
less than two pairs of teeth and not more than three pairs and so on. If n is 1.6,
one pair of teeth are always in contact, whereas two pairs of teeth are in contact
for 60% of the time.
Number of teeth in contact is also given by the contact ratio which is the
ratio of the angle of action to the pitch angle.
4
Theory of Machines
Example 10.4 Two gears in mesh have a module of 8 mm and a pressure angle
of 20°. The larger gear has 57 while the pinion has 23 teeth. If the addenda on
pinion and gear wheels are equal to one module, find
(i) the number of pairs of teeth in contact
(ii) the angle of action of the pinion and the gear wheel
(iii) the ratio of the sliding to rolling velocity at
(a) the beginning of contact
(b) the pitch point
(c) the end of contact.
Solution
j = 20°, T = 57, m = 8 mm, t = 23, addendum = 1 m = 8 mm
8 ¥ 57
mT
=
= 228 mm
2
2
Ra = R + m = 228 + 8 = 236 mm
R=
8 ¥ 23
mt
=
= 92 mm
2
2
ra = r + m = 92 + 8 = 100 mm
r=
(i) n =
Arc of contact
Circular pitch
Path of contact + Path of recess
Ê Path of contact ˆ
1
= Á
˜¯ ¥ p m =
cos j
cos j ¥ p m
Ë
È R 2 - R 2 cos 2 j - R sin j ˘ + È r 2 - r 2 cos 2 j - r sin j ˘
Î a
˚ Î a
˚
=
cos j ¥ p m
È (236) 2 - (228)2 cos 2 20∞ - 228 sin 20∞˘
Î
˚
2
2
2
È
+ Î (100) - (92) cos 20∞ - 92 sin 20∞˘˚
=
cos 20∞ ¥ p ¥ 8
=
20.97 + 18.79
1
= 42.31 ¥
p ¥ 8 = 1.68
cos 20∞ ¥ p ¥ 8
(ii) Angle of action = Angle traversed by the arc of contact
\
dp = Arc of contact ¥
360
2p r
360
= 42.31 ¥ 2
p ¥ 92 = 26.3° or
dg = 42.31 ¥
360
2p ¥ 228
= 10.63° or 10°38¢
26°18¢
Gears
(iii) (a)
Sliding velocity
Rolling velocity
=
=
5
(w p + w g ) ¥ Path of approach
Pitch line velocity (= w p ¥ r )
(w
p
+
)
23
w ¥ 20.97
57 p
w p ¥ 92
= 0.32
(b)
(w p + w p ) ¥ 0
Sliding velocity
=
=0
Rolling velocity
Pitch line velocity
(
)
23
w p + w p ¥ Path of recess
Sliding velocity
57
(c)
=
Rolling velocity
wp ¥ r
=
(1 + 5723 ) ¥ 18.79
92
= 0.287
Example 10.5 Two 20° gears have a module pitch of 4 mm. The number of teeth
on gears 1 and 2 are 40 and 24 respectively. If the gear 2 rotates at 600 rpm,
determine the velocity of sliding when the contact is at the tip of the tooth of gear 2.
Take addendum equal to one module.
Also, find the maximum velocity of sliding.
Solution
Let 1 be the gear wheel and 2 the pinion.
j = 20°, T = 40, Np = 600 mm, t = 24, m = 4 mm
Addendum = 1 module = 4 mm
4 ¥ 40
mT
=
= 80 mm
2
2
Ra = 80 + 4 = 84 mm
R=
4 ¥ 24
mt
=
= 48 mm
2
2
ra = 48 + 4 = 52 mm
r=
t
24
= 600 ¥
= 360 rpm
T
40
(i) Let pinion (gear 2) be the driver.
The tip of the driving wheel is in contact with a tooth of the driven wheel at the
end of engagement. Thus, it is required to find the path of recess which is
obtained from the dimensions of the driving wheel.
Ng = Np ¥
Path of recess =
=
ra2 - (r cos j )2 - r sin j
(52) 2 - (48 cos 20∞)2 - 48 sin 20∞
= 9.458 mm
6
Theory of Machines
Velocity of sliding = (wp + wg) ¥ Path of recess
= 2p (Np +Ng) ¥ 9.458
= 2p (600 +360) ¥ 9.458
= 57 049 mm/min = 950.8 mm/s
(ii) In case the gear wheel is the driver, the tip of the pinion will be in contact with
the flank of a tooth of the gear wheel at the beginning of contact. Thus, it is
required to find the distance of the point of contact from the pitch point, i.e. path
of approach.
The path of approach is found from the dimensions of the driven wheel which
is again pinion.
Thus, path of approach =
ra2 - (r cos j )2 - r sin j
= 9.458 mm, as before
and velocity of sliding = 950.8 mm/s
Thus, it is immaterial whether the driver is the gear wheel or the pinion, the
velocity of sliding is the same when the contact is at the tip of the pinion.
Maximum velocity of sliding will depend upon the larger path considering any
of the wheels to be the driver.
Consider pinion to be the driver.
Path of recess = 9.458 mm
Path of approach =
=
Ra2 - ( R cos j ) 2 - R sin j
(84) 2 - (80 cos 20∞) 2 - 80 sin 20∞ = 10.117 mm
This is also the path of recess if the wheel becomes the driver.
Maximum velocity of sliding = (wp + wg) ¥ Maximum path
= 2p (600 + 360) ¥ 10.117
= 61024 mm/min = 1017.1 mm/s